A) \[\frac{x}{y}\]
B) \[\frac{{{x}^{2}}}{{{y}^{2}}}\]
C) \[\frac{x}{{{y}^{2}}}\]
D) \[x{{y}^{2}}\]
E) \[\frac{x-1}{{{y}^{2}}}\]
Correct Answer: C
Solution :
\[\because \] \[y=1+x+{{x}^{2}}+....\infty \] \[\therefore \] \[y=\frac{1}{1-x}={{(1-x)}^{-1}}\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{1}{{{(1-x)}^{2}}}(-1)=\frac{1}{{{(1-x)}^{2}}}\] \[\therefore \] \[\frac{dy}{dx}-y=\frac{1}{{{(1-x)}^{2}}}-\frac{1}{1-x}=\frac{1-1+x}{{{(1-x)}^{2}}}\] \[=\frac{x}{{{(1-x)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}-y=\frac{x}{{{y}^{2}}}\]
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